A linear ODE

To find the eigenvectors, we solve for the roots of the characteristic polynomial:
$$(1-\lambda)(\lambda^2-1) - (\lambda - 1) + (\lambda - 1) = 0 \implies \lambda_1 = \lambda_2 = 1, \lambda_3 = -1$$.

Solving for the eigenvector associated to $\lambda_1 = \lambda_2 = 1$ yields $v_1 = (2, -1, 1)$. Solving for the eigenvector associated to $\lambda_3 = -1$ yields $v_3 = (0, -1, 1)$. Now to find the missing generalized eigenvector one must solve $(A - I)v_2 = v_1$. A solution of this equation is given by $v_2 = (-1, 2, 0)$. Thus the general solution is
$$c_1 v_1 e^t + c_2(v_1 te^t + v_2 e^t) + c_3v_3e^{-t}$$
where $c_i$ are constants.

For the solution where $y(0) = (1, -1, 1)$, one notes that $y(0) = (v_1 + v_3)/2$. Hence $c_1 = 1/2$, $c_2 = 0$ and $c_3 = 1/2$.